Engineering Tech-III
Electric Vehicle Design-04
Activity-04 Table of Contents
4-1.
What is Energy?
4-2.
What is Potential Energy?
4-3.
What is Kinetic Energy?
4-4.
What is the Total Energy of a System?
4-5.
Energy in the Workplace
4-6.
Inertia & Rotational Kinetic Energy
4-7.
Common Shapes and Formulas
4-8.
What is Power?
4-9.
How is Electrical Power Calculated?
4-10.
Force Transformers
4-11.
Ideal & Actual Mechanical Advantage
4-12.
Formulas for Gear-Drive Systems
4-13.
Primary Concepts – Review
4-14.
Unit Conversion Tables
4-15.
Questions
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4-1. What is Energy?
For our purpose, energy is defined as the ability to do work. If an
electric motor is supplied with energy, it can do work. The energy of an
electric motor can be used to turn a pulley that causes a wheel to move.
Sometimes when you use energy to do work, the work results in a different
form of energy. For example, when a pump lifts water from a lake to a
water tower, work is done by the pump. The work is stored in the water
that‟s been moved to a higher position. This stored energy can be used later
to do work. When the water in the tower runs downhill, the energy that was
stored in it can cause sprinklers to rotate or turbines to turn - thereby doing
work.
4-2. What is Potential Energy?
If a force is applied and work is done to change an object‟s vertical
position, the object gains potential energy. It‟s called "potential" because
the energy is "stored" and can be used at a later time. Gravitational potential
energy can be stored in an object by lifting it to a higher position, such as
water in the water tower. Work has to be done to get the water into the
water tower. The gravity will then do the work when the water flows back
down. Also, potential energy can be stored in fluid systems when gases are
compressed (such as air in a tire). And springs that have been stretched or
compressed have potential energy stored in them.
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Potential energy can be stored in electrical systems when positive and
negative charges are separated from one another. In both batteries and
capacitors, electrons are moved.
They change their positions and their
potential energy.
4-3. What is Kinetic Energy?
When an object moves, it demonstrates kinetic energy. However,
work must be done to get the object to move. The work done shows up as
energy - in the form of motion. Kinetic energy does work through energy of
motion. This energy of motion in machines turns shafts, drives screws,
sands surfaces, stirs liquids and changes the shape of materials.
4-4. What is the Total Energy of a System?
Potential energy and kinetic energy are two common energy forms
within a system. But these two energy forms may not add up to the total
energy. To find the total energy you add the potential and kinetic energy to
the heat energy that comes from friction as the system operates.
Since all systems resist movement, some of a system‟s total energy is
always used up to overcome resistance. This energy reappears as heat in an
operating system.
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The law of conservation of energy tells us that a system‟s total energy
stays constant. The law says, "Energy can be neither created nor destroyed."
But energy can change forms.
Think about the pendulum in the grandfather clock. The pendulum
swings back and forth. This action moves gears that indicate time. What‟s
the total energy stored in the pendulum as it swings back and forth?
When the pendulum is at it uppermost travel, the total energy is all
potential energy. When the pendulum is at its vertical position, a position
where it travels the fastest, the total energy is all kinetic energy. When the
pendulum is in any other position, the total energy is a combination potential
and kinetic (and heat energy).
Total Energy = Potential Energy + Kinetic Energy + Heat Energy
ETOT = EP + EK + Heat Losses
Potential Energy = Weight x Height (above Reference Point)
Ep = w x h
Kinetic Energy = ½ x Mass x Speed Squared of the Moving Mass
EK = ½ x m x v2
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4-5. Energy in the Workplace
Locating sources of energy, harnessing energy, and using energy are
major concerns for any nation.
Transportation by air, land and sea is
possible only if energy is available.
The industries of North America
consume large amounts of electrical and thermal energy to manufacture and
market the machinery and goods. For example, automobiles, construction
equipment, factories, airfields, hospitals, department stores, machine shops,
supermarkets, farms and dairies all use energy. If energy weren‟t available,
none of these could exist.
As an example, let‟s look a concrete-breaking machine that has a
hammer weighing in at approximately 1,000 pounds. The hammer is lifted
to a height of four feet and then falls and breaks the concrete. Find the
potential energy stored in the hammer when it is four feet above the ground.
Potential Energy = Weight x Height (above Reference Point)
Ep = w x h
EP = 1,000 pounds x 4 feet
EP = 4,000 foot-lbs
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Let‟s look at another example. This time an electric vehicle that
weighs 250 pounds moves with a speed of 44 feet per second just before it
slams into a curb. Find the kinetic energy of the car when it hits the curb.
Kinetic Energy = ½ * Mass * (Speed of the Moving Mass)2
EK = ½ * m * v2
EK = ½ * 250 pounds * (44)2 feet per second
EK = ½ * 250 pounds * 1,936 fps
EK = ½ * 484,000 ft-lb
EK = 242,000 ft-lb per second
=>
242,000 ft-lb
4-6. Inertia & Rotational Kinetic Energy
The moment of inertia (I) in a rotational mechanical system is similar
to mass in a linear mechanical system. The larger the moment of inertia of
the rotating object, the larger the torque needed to spin it. The moment of
inertia depends on the location of the axis about which the body spins, how
much mass the object has, and how that mass is distributed around the axis
of spin.
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Linear kinetic energy is defined as EK = ½ x m x v2. In rotational
motion, the equation for kinetic energy looks almost the same.
But in
rotational kinetic energy, you substitute the moment of inertia (I) for the
mass (m). And rotational speed (ω) replaces linear speed (v). Thus the
kinetic energy equation for a rotational mechanical system is given by the
equation:
EK = ½ x I x ω2
4-7. Common Shapes and Formulas
Axis
Mass
Particle in a circular orbit
I=mxr
radius
2
Slender rod, axis through center
Axis
ℓ
2
I = 1 / 12 x m x ℓ
Angular ring, cylinder
r1
r2
I = ½ x m x (r12 + r22)
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Solid cylinder
I = ½ x m x r2 + 1/12 x m x ℓ2
ℓ
Thin walled hollow cylinder (wheel)
I = m x r2
Now let‟s find the total rotational kinetic energy of a 20-inch diameter
wheel weighing five pounds, traveling at 32.33 miles per hour or
approximately 544 rpm.
I = m x r2
I = 5 pounds * (544 rpm / 60 seconds-minute)2
I = 5 lbs * (9.07)2 fps
I = 5 lbs * 82.2 fps
I = 411 ft-lbs per second
=>
411 ft-lbs
Now let‟s substitute "I" in the equation EK = ½ * I * ω2
EK = ½ * I * ω2
EK = ½ * 411 ft-lbs * ω2
EK = ½ * 411 ft-lbs * (544 rev/min x 1 min/60 sec x 6.28 rad/rev)2
EK = ½ * 411 ft-lbs * (56.939 rad/sec)2
EK = ½ * 411 ft-lbs * 3,242.012 rad2/sec2
EK = 1,332,538 ft-lb
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Note: A 5-pound, 20-inch diameter wheel turning 544 rpm develops
411 feet per pound of kinetic energy. This is also the energy required to
rotate the 20-inch wheel at a constant rate of 544 rpm.
4-8. What is Power?
Power is the rate at which work is performed. We can measure the
power of a device or machine by finding how much work it does in a certain
time interval. Or, by finding how much energy it uses in a certain time.
Earlier in this activity, you learned the difference between linear and
rotational work.
Those concepts both apply to power.
equations for determining each:
Linear Mechanical
Power = Work / Time
P=W/t
P = (F x D) / t
P=Fxv
Rotational Mechanical
Power = Work / Time
P = (T x θ) / t
P=Txω
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Engineering Tech-III
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Let‟s look at our first example of power. A hydraulic cylinder lifts a
steel casting in a manufacturing plant. The casting weighs 200 pounds and
is raised three feet in four seconds at a constant speed. Find the horsepower
of the cylinder.
Power = Work / Time
P=W/t
P = (F x D) / t
P = (200 pounds x 3 feet) / 4 seconds
P = 600 foot-pounds / 4 seconds
P = 150 foot-lbs/sec
-- then -P = 150 ft-lb/sec x (1 horsepower / 550 ft-lb/sec)
P = 0.27 hp
Let‟s look at another example of power. An electric motor develops a
shaft torque of five pounds per foot when the shaft rotates at 3,200 rpm.
Determine the horsepower delivered by the shaft.
Power = Work / Time
P = (T x θ) / t
P=Txω
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P = 5 foot-pounds x (3,200 rpm x 1 min/60 sec x 6.28 rad/rev)
P = 5 ft-lbs x 334.933 rad/sec
P = 1,674.665 ft-lbs/sec
-- then -P = 1,674.665 ft-lb/sec x (1 horsepower / 550 ft-lb/sec)
P = 3.045 hp
4-9. How is Electrical Power Calculated?
Power = Work / Time
P = (V x q) / t
P=VxI
or
P = I2 x R
or
P = V2 / R
Now let‟s look at calculating electrical power. Given an electric
motor that produces an output shaft power of 2 horsepower. While doing
this, it operates on a voltage of 24 volts and draws a current of 75 amps.
Find input power and then determine the efficiency of the motor.
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Input Power
P=VxI
PIN = 24 volts x 75 amps
PIN = 1,800 volt-amps
PIN = 1,800 watts
Efficiency
POUT = 2 horsepower x (746 watts / 1 horsepower)
POUT = 1,492 watts
Efficiency = Power (out) / Power (in) x 100%
Eff = (POUT / PIN) x 100%
Eff = (1,492 watts / 1,800 watts) x 100%
Eff = 0.829 x 100%
Efficiency = 82.9%
Let‟s look at another example. This time you have a 24-volt electric
motor with an output power of 1.0 horsepower and is rated at 88 percent
efficient. Find the current required to run the motor and the amperage draw
of the motor.
Eff = POUT / PIN
PIN = POUT / Efficiency
PIN = 746 watts / 0.88
PIN = 847.7 watts
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--then-PIN = V x I
I = PIN / V
I = 847.7 watts / 24 volts
I = 35.29 amps
4-10.
Force Transformers
Force transformers are simple machines or devices in mechanical,
fluid, and electrical energy systems that change input values of force,
movement, or rate into different output values. Force transformers are never
100 percent efficient; however, some electrical devices approach that value.
Below is a formula for an ideal force transformer:
ForceIN x DistanceIN = ForceOUT x DistanceOUT
FI x DI = FO x DO
4-11. Ideal & Actual Mechanical Advantage
If you apply a force of 100 pounds to a pulley system and lift 200
pounds, the system has an actual mechanical advantage of 2. However, the
force transformer gains force at the expense of distance. You pulled in a
total distance of 10 feet of rope and the pulley system lifted the load a total
of 5 feet.
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The ideal mechanical advantage in this case is also 2. If you divide 2
by 2, the answer is 1. The forces cancel out the distances and the system is
100 percent efficient.
Ideal Mechanical Advantage = DistanceIN / DistanceOUT
IMA = DI / DO
--and--
Actual Mechanical Advantage = ForceOUT / ForceIN
AMA= FO / FI
Let‟s look at an example to see how this works. Dan operates a
wrecker service. The winch cable on his wrecker is rated at 10 tons. To
draw in a disabled truck back onto the road, Dan decides to rig the cable into
a block and tackle. The winch winds in 300 feet of cable while applying a
force of 10 tons. At the same time, the wrecked truck is pulled a distance of
150 feet back onto the road.
Now, calculate the work done by the winch. Then find the force
applied to the disabled truck and find the work done on the disabled truck.
Determine the mechanical advantage of the block and tackle. Assume 100
percent efficiency. Lastly, determine the efficiency of the system.
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Work done by the winch
WorkIN = FIN x DIN
WIN = 20,000 pounds x 300 feet
WIN = 6,000,000 foot-pounds
Force applied to the disabled truck
ForceIN x DistanceIN = ForceOUT x DistanceOUT
FI x DI = FO x DO
FI x DI = FO x DO
FO = (FI x DI) / DO
FO = (20,000 pounds x 300 feet) / 150 feet
FO = 6,000,000 foot-pounds / 150 feet
FO = 40,000 pounds
Work done on the truck
ForceIN x DistanceIN = ForceOUT x DistanceOUT
FI x DI = FO x DO
FI x DI = FO x DO
FO x DO = FI x DI
WO = FI x DI
WO = 40,000 pounds x 150 feet
WO = 6,000,000 foot-pounds
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Ideal Mechanical Advantage (IMA)
Ideal Mechanical Advantage = DistanceIN / DistanceOUT
IMA = DI / DO
IMA = 300 feet / 150 feet
IMA = 2
Actual Mechanical Advantage (AMA)
Actual Mechanical Advantage = ForceOUT / ForceIN
AMA= FO / FI
AMA = 40,000 pounds / 20,000 pounds
AMA = 2
Efficiency
Efficiency = Actual Mechanical Advantage / Ideal Mechanical Advantage
Eff = (AMA / IMA) x 100%
Efficiency = (2 / 2) x 100%
Efficiency = 1 x 100%
Efficiency = 100%
Now let‟s look at levers as a means of providing linear force
transformation.
All levers are used for the same purpose - to gain a
mechanical advantage. The three classes of levers are illustrated in the next
graphic (see top of next page).
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Fout
1st Class
Load
Effort
Fout
Pivot
Fin
Fin
2nd Class
Load
Effort
Fout
Pivot
3rd Class
Fin
Load
Effort
Pivot
What is the relationship between input work, output work and torque
transformers? The force transformer can either increase output force or
output displacement but not at the same time.
WorkIN = WorkOUT
TI x θI= TO x θO
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4-12. Formulas for Gear-Drive Systems
Definition of Symbols

rI and rO = radius of input and output gears

ωI and ωO = angular speed of input and output gears

NI and NO = total number of teeth on input and output gears.
IMA = rO / rI
(IMA equals the ratio of output to input radii)
IMA = NO / NI
(IMA equals the ratio of output to input total teeth count)
IMA = ωI / ωO
(IMA equals the ratio of input to output angular speeds)
IMA = TO / TI
(TMAequals the „ratio of output to input torques, no friction)
ωI / ωO = rO / rI
(Angular speeds and gear radii are inversely related)
ωI / ωO = NO / NI
Angular speeds and gear- teeth count are inversely related)
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Now let‟s look at an example to see how this actually works. The
input gear in the gearbox on your electric vehicle has 16 teeth. The output
gear has 74 teeth. The output speed is 500 rpm. This causes the car to move
forward at 33 miles per hour. Calculate the ideal mechanical advantage
(IMA) of the transmission. Determine the rotational speed (ωI) of the input
gear.
Mechanical Advantage
IMA = NO / NI
IMA = 74 teeth / 16 teeth
IMA = 4.625
Rotational Speed of Input Gear
IMA = ωO / ωI
IMA = ωI / 500 rpm
4.625 = ωI / 500 rpm
4.625 x 500 rpm = ωI
ωI = 2,312.5 rpm
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4-13. Primary Concepts – Review
PHYSICAL SCIENCE: Understand structures and properties of
matter and changes that occur in the physical world.
MOTION: Understand fundamental forces, their forms, and their
effects on motion. Describe and explain the effects of multiple forces acting
on an object.
ENERGY: Understand the interactions of energy and matter.
Compare and contrast form and behaviors of various types of energy.
Describe and explain a variety of energy transfers and transformations.
SCIENCE AND TECHNOLOGY: Understand the interconnections
among science, technology, and society.
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14-4. Unit Conversion Tables
Unit Conversion Tables – Length
Convert from
To
Equation
Meters
Feet
meters x 3.28
Feet
Meters
feet x 0.305
Kilometers
Miles
Kilometers x 0.621
Miles
Kilometers
Miles x 1.609
Unit Conversion Tables – Area
Convert from
To
Square meters (m2)
Square feet
Square feet (ft2)
Square meters
Equation
m2 x 10.76
ft2 x 0.093
Unit Conversion Tables – Power
Convert from
To
Equation
Kilowatts
Watts
Kilowatts x 1000
Watts
Kilowatts
Watts x 0.001
Megawatts
Kilowatts
Megawatts x 1000
Kilowatts
Megawatts
Kilowatts x 0.001
Watts
BTU/hr
Watts x 3.413
BTU/hr
Watts
BTU/hr x 0.293
Kilowatts
BTU/hr
Kilowatts x 3414
BTU/hr
Kilowatts
BTU/hr x 0.000293
Watts
Horsepower
Watts x 0.00134
Horsepower
Watts
Horsepower x 746
Kilowatts
Horsepower
Kilowatts x 1.34
Horsepower
Kilowatts
Horsepower x 0.746
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Unit Conversion Tables – Speed
Convert from
To
Meters per second (mps)
Miles per hour (mph)
Miles per hour (mph)
Meters per second (mps)
Knots
Miles per hour (mph)
Meters per second (mps) Kilometers per hour (kph)
Equation
mps x 2.24
mph x 0.447
knots x 1.15
mps x 3.6
Unit Conversion Tables – Energy
Convert from
To
Equation
Watt-hours
Kilowatt-hours
Watt-hours x 0.001
Kilowatt-hours
Megawatt-hours
Kilowatt-hours x 0.001
Watt-hours
BTU
Watt-hours x 3.413
BTU
Watt-hours
BTU x 0.293
Kilowatt-hours
BTU
Kilowatt-hours x 3414
BTU
Kilowatt-hours
BTU x 0.000293
Watt-hours
Horsepower-hours
Watt-hours x 0.00134
Horsepower-hours
Watt-hours
Horsepower-hour x 746
Kilowatt-hours
Kilowatt-hours
Kilowatt-hours x 1.34
Horsepower-hours
Horsepower-hours
Horsepower-hr x 0.746
4-15. Questions
1.
Given an electric vehicle that weighs 450 pounds moving with a
speed of 40 feet per second, find the kinetic energy of the car as it
travels.
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2.
Electric Vehicle Design-04
Calculate the total rotational kinetic energy of a 20-inch diameter
wheel weighing 6.2 pounds, traveling at 37.95 miles per hour or
approximately 544 rpm.
3.
An electric motor develops a shaft torque of 4.6 pounds per foot when
the shaft rotates at 3,600 rpm. Calculate the power “P” and determine
the horsepower “hp” delivered by the shaft.
4.
An electric motor develops a shaft torque of 11.1 pounds per foot
when the shaft rotates at 2,700 rpm. Calculate the power “P” and
determine the horsepower “hp” delivered by the shaft.
5.
Given an electric motor that produces an output shaft power of 3.2
horsepower. While doing this, it operates on a voltage of 24 volts and
draws a current of 105 amps. Find input power and then determine
the efficiency of the motor.
6.
Given an electric motor that produces an output shaft power of 4.6
horsepower. While doing this, it operates on a voltage of 36 volts and
draws a current of 125 amps. Find input power and then determine
the efficiency of the motor.
7.
Given a 24-volt electric motor with an output power of 2.75
horsepower and is rated at 84 percent efficient. Find the current
required to run the motor and the amperage draw of the motor.
8.
Given a 36-volt electric motor with an output power of 5.5
horsepower and is rated at 86 percent efficient. Find the current
required to run the motor and the amperage draw of the motor.
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9.
Electric Vehicle Design-04
The input gear in the gearbox on your electric vehicle has 18 teeth.
The output gear has 60 teeth. The output speed is 1,250 rpm. This
causes the car to move forward at 28 miles per hour. Calculate the
ideal mechanical advantage (IMA) of the transmission. Determine
the rotational speed (ωI) of the input gear.
10.
The input gear in the gearbox on your electric vehicle has 24 teeth.
The output gear has 60 teeth. The output speed is 1,050 rpm. This
causes the car to move forward at 37 miles per hour. Calculate the
ideal mechanical advantage (IMA) of the transmission. Determine
the rotational speed (ωI) of the input gear.
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